LAWS OF STOICHIOMETRY: DESCRIPTION, EXAMPLES AND EXERCISES - CHEMISTRY - 2024

The laws of stoichiometry describe the composition of the different substances, based on the relationships (by mass) between each species involved in the reaction.

All existing matter is formed by the combination, in different proportions, of the different chemical elements that make up the periodic table. These unions are governed by certain laws of combination known as the laws of stoichiometry or weight laws of chemistry.

These principles are a fundamental part of quantitative chemistry, being indispensable for balancing equations and for operations as important as determining which reactants are needed to produce a specific reaction or calculating how many of these reactants are needed to obtain the expected quantity of products..

The four laws are widely known in the chemical field of science: the law of conservation of mass, the law of definite proportions, the law of multiple proportions and the law of reciprocal proportions.

The 4 Laws of Stoichiometry

When you want to determine how two elements combine through a chemical reaction, the four laws described below must be taken into account.

Law of Conservation of Mass (or "Law of Conservation of Matter")

It is based on the principle that matter cannot be created or destroyed, that is, it can only be transformed.

This means that for an adiabatic system (where there is no transfer of mass or energy from or to the surroundings) the amount of matter present must remain constant over time.

For example, in the formation of water from gaseous oxygen and hydrogen, it is observed that there is the same number of moles of each element before and after the reaction, so the total amount of matter is conserved.

2H ₂ (g) + O ₂ (g) → 2H ₂ O (l)

Exercise:

Q.- Show that the previous reaction complies with the law of conservation of mass.

A.- First, we have the molar masses of the reactants: H ₂ = 2 g, O ₂ = 32 g and H ₂ O = 18 g.

Then, add the mass of each element on each side of the reaction (balanced), resulting in: 2H ₂ + O ₂ = (4 + 32) g = 36 g on the reactant side and 2H ₂ O = 36 g on the products side. Thus it has been shown that the equation complies with the mentioned law.

Law of definite proportions (or “Law of constant proportions”)

It is based on the fact that each chemical substance is formed from the combination of its constituent elements in defined or fixed mass relationships, which are unique for each compound.

The example of water is given, whose composition in its pure state will invariably be 1 mole of O ₂ (32g) and 2 moles of H ₂ (4g). If the greatest common divisor is applied, it is found that one mole of H ₂ reacts for every 8 moles of O ₂ or, which is the same, they combine in a ratio of 1: 8.

Exercise:

Q.- You have one mole of hydrochloric acid (HCl) and you want to know what percentage each of its components is in.

A.- It is known that the union ratio of these elements in this species is 1: 1. And the molar mass of the compound is about 36.45 g. Similarly, the molar mass of chlorine is known to be 35.45 g and that of hydrogen is 1 g.

To calculate the percentage composition of each element, the molar mass of the element (multiplied by its number of moles in one mole of the compound) is divided by the mass of the compound and this result is multiplied by one hundred.

Thus:% H = x 100 = 2.74%

y% Cl = x 100 = 97.26%

From this it follows that, regardless of where the HCl comes from, in its pure state it will always be made up of 2.74% hydrogen and 97.26% chlorine.

Law of multiple proportions

According to this law, if there is a combination between two elements to generate more than one compound, then the mass of one of the elements joins with an invariable mass of the other, preserving a relationship that is manifested through small integers.

Carbon dioxide and carbon monoxide are given as examples, which are two substances made up of the same elements, but in dioxide they are related as O / C = 2: 1 (for each C atom there are two O's) and in the monoxide its ratio is 1: 1.

Exercise:

Q.- There are five different oxides that can be produced in a stable way by combining oxygen and nitrogen (N ₂ O, NO, N ₂ O ₃, N ₂ O ₄ and N ₂ O ₅).

A.- It is observed that the oxygen in each compound is increasing, and that with a fixed proportion of nitrogen (28 g) there is a ratio of 16, 32 (16 × 2), 48 (16 × 3), 64 (16 × 4) and 80 (16 × 5) g of oxygen respectively; that is, we have a simple ratio of 1, 2, 3, 4 and 5 parts.

Law of reciprocal proportions (or "Law of equivalent proportions")

It is based on the relationship between the proportions in which an element is combined in different compounds with different elements.

In other words, if a species A joins a species B, but A also combines with C; It follows that if elements B and C are joined, their mass ratio corresponds to the masses of each one when they join in particular with a fixed mass of element A.

Exercise:

Q.- If you have 12g of C and 64g of S to form CS ₂, you also have 12g of C and 32g of O to produce CO ₂ and finally 10g of S and 10g of O to produce SO ₂. How can the principle of equivalent proportions be illustrated?

A.- The proportion of the masses of sulfur and oxygen in combination with a defined mass of carbon is equal to 64:32, that is, 2: 1. So, the ratio of sulfur and oxygen is 10:10 when joining directly or, what is the same, 1: 1. So the two relations are simple multiples of each species.

References

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